Transient Analysis of RC Circuit with Multiple Voltage Sources

To solve the given problem, let us analyze the circuit step-by-step for $t < t_0$ (key $K$ open) and for $t \ge t_0$ (key $K$ closed).

1. Circuit Analysis for $t < t_0$ (Key $K$ Open)

Let the right-hand vertical wire be at a reference potential of $0\text{ V}$. Let $V_L$ be the potential of the left-hand vertical node.

The batteries are oriented such that their negative terminals are on the left and positive terminals are on the right.

The currents leaving the left node $L$ through the top three branches are:

• Branch 1 (Top branch with $15\text{ V}$ battery and resistance $R$): $I_{\text{top}} = \frac{V_L - (-15)}{R} = \frac{V_L + 15}{R}$

• Branch 2 (Middle branch with $5\text{ V}$ battery and $1\ \Omega$ resistor): $I_{1\Omega} = \frac{V_L - (-5)}{1} = V_L + 5$

• Branch 3 (Middle branch with $3\ \Omega$ resistor): $I_{3\Omega} = \frac{V_L - 0}{3} = \frac{V_L}{3}$

The current $I_1$ defined in the diagram flows to the left through the $3\ \Omega$ resistor, so: $I_1 = -I_{3\Omega} = -\frac{V_L}{3}$

We are given that a current of $1\text{ A}$ flows through the $1\ \Omega$ resistor, which means: $|I_{1\Omega}| = |V_L + 5| = 1\text{ A}$

This gives two potential values for $V_L$:

1. $V_L + 5 = 1 \implies V_L = -4\text{ V} \implies I_1 = \frac{4}{3}\text{ A}$
2. $V_L + 5 = -1 \implies V_L = -6\text{ V} \implies I_1 = 2\text{ A}$

From Option (B), $I_1 = 2\text{ A}$, which uniquely gives: $V_L = -6\text{ V}$

Thus, Option (B) is correct.

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2. Determination of Resistance $R$

Applying Kirchhoff's Current Law (KCL) at node $L$ for $t < t_0$: $I_{\text{top}} + I_{1\Omega} + I_{3\Omega} = 0$

$\frac{V_L + 15}{R} + (V_L + 5) + \frac{V_L}{3} = 0$

Substituting $V_L = -6\text{ V}$: $\frac{-6 + 15}{R} + (-6 + 5) + \frac{-6}{3} = 0$

$\frac{9}{R} - 1 - 2 = 0 \implies \frac{9}{R} = 3 \implies R = 3\ \Omega$

Thus, Option (A) is correct.

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3. Circuit Analysis for $t \ge t_0$ (Key $K$ Closed)

When key $K$ is closed at $t = t_0$, the fourth branch containing the capacitor $C = 2\ \mu\text{F}$ and the $3\ \Omega$ resistor is connected to the circuit.

To analyze the charging process, we determine the Thevenin equivalent circuit connected across the capacitor branch:

1. Thevenin Voltage ($V_{\text{th}}$): This is the open-circuit potential difference across the left and right nodes before connecting the 4th branch: $V_{\text{th}} = |V_L| = 6\text{ V}$

2. Thevenin Resistance ($R_{\text{eq}}$): The equivalent resistance of branches 1, 2, and 3 connected in parallel: $\frac{1}{R_{\text{eq}}} = \frac{1}{R} + \frac{1}{1} + \frac{1}{3} = \frac{1}{3} + 1 + \frac{1}{3} = \frac{5}{3}\ \Omega^{-1} \implies R_{\text{eq}} = 0.6\ \Omega$

3. Total Resistance of the RC Loop ($R_{\text{total}}$): $R_{\text{total}} = R_{\text{eq}} + 3\ \Omega = 0.6\ \Omega + 3\ \Omega = 3.6\ \Omega$

4. Time Constant ($\tau$): $\tau = R_{\text{total}} \times C = 3.6\ \Omega \times 2\ \mu\text{F} = 7.2\ \mu\text{s}$

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4. Evaluation of Current at $t = t_0 + 7.2\ \mu\text{s}$

Since the capacitor is initially uncharged ($V_C(t_0^+) = 0$), the initial current in the capacitor at $t = t_0^+$ is: $I_C(t_0^+) = \frac{V_{\text{th}}}{R_{\text{total}}} = \frac{6\text{ V}}{3.6\ \Omega} = \frac{5}{3}\text{ A}$

The current as a function of time for $t \ge t_0$ is: $I_C(t) = I_C(t_0^+) e^{-\frac{t - t_0}{\tau}} = \frac{5}{3} e^{-\frac{t - t_0}{\tau}}\text{ A}$

At $t = t_0 + 7.2\ \mu\text{s} = t_0 + \tau$: $I_C(t_0 + \tau) = \frac{5}{3} e^{-1} = \frac{5}{3} \times 0.36 = 0.60\text{ A}$

Thus, Option (C) is correct.

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5. Evaluation of Final Charge ($t \rightarrow \infty$)

As $t \rightarrow \infty$, the capacitor is fully charged and acts as an open circuit ($I_C(\infty) = 0$). The potential difference across the capacitor reaches $V_{\text{th}} = 6\text{ V}$.

The steady-state charge on the capacitor is: $Q(\infty) = C \cdot V_{\text{th}} = 2\ \mu\text{F} \times 6\text{ V} = 12\ \mu\text{C}$

Thus, Option (D) is correct.

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Conclusion

All statements (A), (B), (C), and (D) are correct.